Find the magnitude and direction angle of r=-48i+6j.
I think I understand how to get the magnitude but not understanding the direction angle...
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OpenStudy (anonymous):
So to get the magnitude would be
\[\sqrt{-48^{2}+6^{2}}\]
OpenStudy (anonymous):
I think.
OpenStudy (anonymous):
|dw:1360695998634:dw|
OpenStudy (anonymous):
you are right on the magnitude
OpenStudy (anonymous):
So the direction angle goes back to a unit circle?
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OpenStudy (anonymous):
Or am I just crazy
OpenStudy (anonymous):
yeah it pretty much does, all the unit circle is is a way to look at angles, so take
tan(x) = opp/ adj
tan(x) = 6/48 <-- 48 is positive because it's just a magnitude
the angle in the triangle is atan(6/48)
OpenStudy (anonymous):
so the direction from positive x is 180 - that angle found
OpenStudy (anonymous):
Okay then would the magnitude be kept in a fraction form 'cause the decimal goes on forever o.o
OpenStudy (anonymous):
|dw:1360696335478:dw|
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OpenStudy (anonymous):
the magnitude is up to you or however your teacher wants it. it is safe to round off usually to 2 decimals